help solve this simultaneous equation?

Question

3x+4y= 6x
x-3y= -8

Answer 1

1st eqtn:
3x+4y=6x gives 3x-6x+4y=0
so 4y-3x=0 thus 4y=3x and y=3/4 x--'a'
2nd eqtn:
x-3y+8=0
putting value of y from eqtn 'a'
x-9/4x+8=0
so -5/4x+8=0 thus 5/4x=8
which gives x=32/5
putting value of x in 'a' we get the value of y=3/5

Answer 2

3x + 4y = 6x ==> 4y = 3x
x - 3y = -8

Using the second equation, x = 3y - 8. Then, using the simplified version of the first equation and substituting this expression for x, 4y = 3(3y - 8). This is now a problem in one variable, which you should be able to solve. Once you have solved for y, use it in the formula x = 3y - 8 to get the value of x.

Answer 3

from the 2nd equation we get
x=3y-8
substituting the value of x in eqn 1
3(3y-8)+4y=6(3y-8)
9y-24+4y=18y-48
9y+4y-18y= -48+24
-5y= -24
y=-24/-5=4 4/5
and x=3*24/5 -8
=72/5 -8
=32/5=6 2/5

Answer 4

first = -3x + 4y = 0. Multiply second by 3:
3x - 9y = -24, and add
-5y = -24, or y = 24/5
from second, x - 72/5 = -8, or x = 32/5.
so x,y = (32/5, 24/5)

Answer 5

TAKE SECOND EQUATION......

X-3Y= -8
MULTIPLY THE WHOLE EQN. BY 3

SO IT BECOMES

3X-9Y= -24.

NOW TAKE FIRST EQN.
3X+4Y=6X
SO
3X-6X+4Y=0

THAT IS -3X+4Y=0
WRITE BOTH EQNS AS FOLLOWS
-3X+4Y=0
3X-9Y= -24
ON ADDING THE ABOVE TWO EQNS WE GET,
-5Y=-24
5Y=24
SO Y=24/5.
PUT VALUE OF Y IN EQN -3X+4Y=0
SO -3X+4MULTIPLIED BY24/5=0
SO -3X+96/5=0
TAKE L.C.M AND THEN SHIFT DENOMINATOR TO RIGHT SIDE OF EQUAL SIGN WE GET,
-15X+96=0
-15X=-96
15X=96
X=96/15
X=32/5

Answer 6

3x + 4y = 6x
4y = 6x - 3x = 3x (transposing)
y = 3x/4

substituting tht value in the 1st eqn..
x - 3(3x/4) = -8
x - 9x/4 = -8
(4x-9x)/4 = -8
-5x= -32
x=32/5
y=3x/4 = 24/5